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Identificaiton of Beam requirements
2

Identificaiton of Beam requirements

Identificaiton of Beam requirements

(OP)
I am trying to calculate the size of a Universal Beam that I will need to replace an  old wooden beam.

It is a simply supported beam that needs to hold up a uniformly distributed load of around 1.5 tonnes, over a length of 14.5ft. The beam is supported at either end by an upright column.

I have calculated this weight using the following:
Surface Area of Slate & Wood Roof = 103 sq ft @ 16lbs / sq ft
 = 1624 lbs

Surface area of Wood Panal and Partially Glazed wall = 100sq ft @ 6lbs / sq ft =609lbs

TOTAL Weight  = 2233 lbs

Safety Factor = 2233*1.5
= 3349lbs

I  have tried using  BM = wl^2/8 and D= 5/384 x wl^3/EI , but I am not having much luck as I am getting figures that say a 203 x133 will be sufficient but this seems light.

Any help would be greatly appreciated.  If you could show me the calculations it would be fantastic so I can see where I am going wrong.

Rob
 

RE: Identificaiton of Beam requirements

Even the dead loads seem a bit light, but I don't see you using any live loads. Be it interior or exterior application, you use some live load. Since you name slate one would think it is a roof and hence most likely would settle the live load more or less equals at the worst case live load, this with a significant snow load. So look what th snow load needs be at the location, and if too weak, put alternatively anyway some live load for repairs. Almost any code on loads has provision for this.

RE: Identificaiton of Beam requirements

i'd be carefull about the mixed units ... 3349lbs is a running load of 231 lbs/ft = 3.37N/mm. so the moment is calculated to be 3.37*(14.5*12*25.4)^2/8 = 8.2E6Nmm ... is that the capacity of your section ?

RE: Identificaiton of Beam requirements

I assume that a Universal beam is similar to a Wide Flange or I beam.  You have not told us the properties of the beam, so it is difficult to respond intelligently to your question.

If w represents the uniform load per foot, then:

M = wL2/8

D = 5/384 * w*L4/EI (contrary to your formula)

if W represents the total uniform load on the beam, namely wL then:

M = WL/8

D = 5/384 *W * L3/EI

The bending strength of a steel beam depends on the lateral support of the compression (top) flange.  But for such light loading as you are assuming, a section of 200 x 133 would likely be adequate.

The factored maximum moment is 3349 * 14.5/8 = 6070'# (8.22kN-m)

Deflection need not be checked based on experience, but if you are going to check it, keep the units consistent.

According to my steel handbook, W200x19 has a resisting moment of 58.1 kN-m when the top flange is continuously supported and 16.4 kN-m when supported at 5m centers.

W200x19 is 200mm deep, has a flange width of 102mm and weighs 19 kg/m.  Its imperial equivalent is W8x13 which means it has a depth of 8" and weighs 13 pounds per foot.

BA

RE: Identificaiton of Beam requirements

(OP)
HI Everyone, thaks for your input.  Looking back I should have included a little more information.

In my original calculations I applied a 40psi load for snow and wind, but I forgot the live load of individuals in the room itself.

I also used D = 5/384 *W * L3/EI instead of D = 5/384 *W * L4/EI, which would explain why I got a lighter value for the Inertion.

The I-Beam is to replace an old wooden beam, so I am trying to define what size of beam I would require, hence why I did not incldue properties of the beam.

I have carried out my calculations in imperial purely becasue I have an old structural steelwork handbook.  I only stated the 1.5 tonne uniform lode in metric as I assumed that is what everyone would recognise.

If the factored moment is 8.22kN-m without the live loads, If I add in my live loads to get a new BM, then this should give me a more realistic value.  Should I then take that total BM value and select a beam smaller than the W200x19 depending on  ym asnwer?

Thanks again, and apologies for my ignorance, I studied a small amount of structural at Uni, but 15 years on it seems to elude me winky smile !

Rob

RE: Identificaiton of Beam requirements

(OP)
Apologies, "40 PSI" should read "40lbs Sq Ft".

Thanks

RE: Identificaiton of Beam requirements


gems:  What are the cross-sectional dimensions of the existing wood beam?
 

Ralph
Structures Consulting
Northeast USA

RE: Identificaiton of Beam requirements

(OP)
Hi RHTPE

The existing beam is 50mm x 150mm, however ti used to be supported at 7ft centres, and we want to be able to remove the beam, hence the new 14ft span.

Thanks

rob
 

RE: Identificaiton of Beam requirements

Existing wood beam is only one 2x6? Your safety factor? Are you sure about tributary areas and loads?  Why not laminate wood, and make a built up wood beam to avoid trying to get a steel beam in there in the first place.
Looks like a residential DIY. D is ??

RE: Identificaiton of Beam requirements


gems:   Do you really want to try to muscle in a 200+ pound steel beam in amongst the temporary supports you will have to install to support the floor joists (or rafters) until the replacement is complete?

I'm with beton1 - install some temporary supports to take the load off the existing beam and laminate some additional lumber to it.  Perhaps build a flitch beam in-place.

The difficulty with these kinds of questions is that the folks you are asking advice from cannot see what you have seen, and we only get a small piece of the puzzle.  Which leads to additional questions, and by the time we converge on a possible solution, the complexion of the challenge has changed dramatically.

A picture is worth a thousand words ....

Ralph
Structures Consulting
Northeast USA

RE: Identificaiton of Beam requirements

Another idea is to use a channel each side of the existing beam and bolt it through, then remove the central column.  Before doing that, however it would be wise to ensure that the two remaining columns and foundations are capable of safely sustaining the entire load.

BA

RE: Identificaiton of Beam requirements

(OP)
Hi Again

Thanks for all the input, it is greatly appreciated.

RHTPE, good, point a picture would definately help.

The structure we are talking about is  the bit on the right hand side  that looks like a shed on stilts.  Which is basically all it is apart from the fact it has a pitched / slated roof.

The beam is to span from point A - Point B and is  7m long  with the existing beams anchored into the house, but for ease I am looking at these as anchored simply supported beams. (The overall structure is approx 4.5m x 7m)

http://www.qems.biz/daphnia/house/house1.jpg

The house has a great deal of history to it and we are eager to preservce as much of it as we can.  Unfortunately the "Summer house" is completely rotten though years of neglect.  The roof however is perfect.  We have had an engineering / scaffolding company, support the entire roof, and the sub-structure has been removed.  At present the joists and floor are also supported and in place, but this is purely for H&S reasons.

http://www.qems.biz/daphnia/house/house2.jpg

Basically the entire structure is beaing removed and replaced excluding the roof.

Here is why we need to replace the beam!
http://www.qems.biz/daphnia/house/beam.jpg

As you can see the floor is in place and supported fromt the underside.  The existing steel structure that you can see in the picure was a temporary structure put in a few years ago as all wooden supports were completely rotten.  It isn't pretty and over engineered but it was done as an emergency situation.  The Column is a 150mm * 150mm Box Section with a 10 mm wall.

So, now that I have rambled on for some time, as you can see it is a relatively simple idea.  The reason I want to be able to calculate the beam size is, partially curiosity and partially ensuring that I know the minimum requirements when the engineering company inform me of what they are putting in.  We have spent the last 7 years putting our life ansd soul into this place, and I tend to be a bit ptotective when we get work done externally! winky smile

On that note, here are my calcs for Maximum Bending Modulus etc.

Roof Load = 7.66 N/m2 x 31.5 (Surface area)= 24.13 kN
Floor Load = 167.58 N/m2 x 31.5 (Surface Area) = 5.3kN
Walls = 287.28N/m2 x 31.5 (Surface Area) = 9.04kN
Joists = 301.64N/m2 x 5.3 (Surface area of Joists) = 1.5kN
Snow & Live = 700N/m2 x 31.5 (This is excessive but better safe than sorry) = 22 kN
Wind Load = 430.92 N/m2 x 31.5 (Surface Area) = 13.54kN

Total Load  (UDL) = 75.5kN
Load per M = 5.2kN/m

Moment = (75.5*14.5)/8 = 136kN

Shear = 5.2 x 14.5/2 = 37.7kN

The difficulty I am having is transferring this into a beam size.

Thanks




 

RE: Identificaiton of Beam requirements

(OP)
Please ignore the calcs above: I did the schoolby error of getting the units mixed up  (despite the previous warning)!!

Roof Load = 766.08 N/m2 x 9.6 (Surface area)= 7.35 kN
Floor Load = 167.58 N/m2 x 9.6 (Surface Area) = 1.6kN
Walls = 287.28N/m2 x 9.6 (Surface Area) = 2.7kN
Joists = 301.64N/m2 x 5.3 (Surface area of Joists) = 1.6kN
Snow & Live = 700N/m2 x 9.6 (This is excessive but better safe than sorry) = 6.72 kN
Wind Load = 430.92 N/m2 x 9.6 (Surface Area) = 4.136kN

Total Load  (UDL) = 24.11kN
Load per M = 5.36/m

Moment = (24.11*4.5)/8 = 13.56kN

Shear = 5.36 x 4.5/2 = 12.06kN

Cheers

Rob

 

RE: Identificaiton of Beam requirements

(OP)
Apologies again, I should really proof read a little better prior to posting;  the Building size is 4.5m x 2.13 m not "4.5m x 7m" as stated above.

 

RE: Identificaiton of Beam requirements

gems:

Maybe someone has already pointed out, your repeat mistakes are resulted from two no nos:

1. Lack of a simple sheet lists all design parameters, accompanied with a simple sketch indicating dimensions, member sizes.
2. Mixing two different units in statement and cal.

Not criticing, just wish you do well.

RE: Identificaiton of Beam requirements

(OP)
Hi Kslee100

I agree entirely I started using imperial and then referd to Metric.  All is now changed to Metric.

The simple drawings I have, but didn';t think to san and post online.

I assumed I would be able to figure this out from vague memories of uni, obviously not the case.  I will keep scribbling for my own curiosity, and won't take up any more you guy's time.  

Thanks again for the help.

 

RE: Identificaiton of Beam requirements

Hire an engineer.

RE: Identificaiton of Beam requirements

he's just rusty, need little time to get back.

RE: Identificaiton of Beam requirements

(OP)
Very rusty, winky smile As for hiring an engineer, it would kind of defeat the purpose of satisfying my curiosity winky smile.  I will not have the say of what is going in , I just want to have an understanding.

Rob

RE: Identificaiton of Beam requirements

Please keep communication open. It can be frustrated at times because we are not in a face to face situation, things tend to take a while to clear up. But more often than not, there is someone hold the "key" to your question, and sometimes a little criticism/comment can be helpful too. Don't take it personal.  

RE: Identificaiton of Beam requirements

There is still a problem with your calculation.  You stated that the building size is 4.5m x 2.13m.  So the tributary width of floor and roof going to the beam should be 2.13/2 = 1.07m should it not?

When you calculated your roof load and floor load, you assumed the entire area was carried by the beam.

When you considered snow and live load, you did not take into account 1.9 kPa for the floor.

BA

RE: Identificaiton of Beam requirements

(OP)
Hi Kslee1000

Not a problem, it's OK, I wasn't taking it personally, I agree with all of the comments, just a little confused about the "hire and Engineer" one. I am finding it frustrating stumbling at such an early hurdle when I feel I should be able to remember this (even if it was 15 years agowinky smile ). SUggesting that I hire an engineer implies that  I should not be asking this question on the forum.

That said, I have taken on board your suggestions and tried to do this with a  more methodology and structure, please see below:

Roof :
Total surface area:  4.5*1.6=7.2m^2
SA attributed to beam:  7.2m^2/2 = 3.6m^2

Floor:
Total surface area:  4.5*2.13  = 9.6m^2
SA attributed to beam: 9.6m^2/2 = 4.8m^2
Wall:
Total Surface Area:  = 9.6m^2

Joists = 5.3m^2 / 2 = 2.65m^2

DEAD LOADS
Roof : 766.08N/m^2 8 3.6m^2 = 2.758kN
Floor:  167.58N/m^2 * 4.8m^2 = 0.85kN
Wall : 287.287 N/m^2 * 9.6m^2 = 2.75kN
Joists: 301.64 * 2.65m^2 = 0.8kN

LIVE LOADS
Wind: 430.92N/m2 * 3.6m^2 = 1.55kN
Snow: 700N/m^2 * 3.6m^2 = 2.52kN
Floor Live Load: 1900N/m^2 * 4.8 = 9.12kN (I am not sure what this is but I have added it  under your instruction – I am assuming it is for furniture / people etc)

Total Deal Load = 7.133kN
Total Live Load = 13.19kN

Dead Load Safety Factor = 1.4
Live Load Safety Factor = 1.6

Total UDL = 1.4*7.113kN + 1.6 * 13.19kN
=31.104kN

=6.912kNm for the span of 4.5 m.

Maximum Moment
M=wl/8 (where w is the total UDL)
M= (31.104 * 4.5) /8 = 17.45kNm

Maximum Shear
Shear = (6.912kN * 4.5)/2 = 15.55kN

Moment Capacity for a fully restrained beam is equal to pySx so the Sx I  require is 17.45*10^3/265 (py=265N/mm^2)  ( I am not sure if I am using the correct " strength value)

Sx required = 65.84 mm^3.

I now need to select a beam that has an Sx greater than 65.84mm^3.
 
Now to check the Deflection
Maximum Deflection must not be greater than span/250  = 4500/250= 18mm

D = 5/384 *W * L3/EI
I= (5*(31.104)*4.5^3)/(384*205*18) *10^5
I = (14171.76 / 1416960) *10^5 = 1000.15mm^4

So technically if I choose a section which has an I-value greater than 1000 mm^4 and check that its Sx > 65.84mm^3  then this should be the beam for the job. (assuming I have not made any more school-boy errors)

The parts I am unsure of are:

If I am using the correct Youngs Modulus value, and the correct strength value.

If I use the UDL with the safety factor when calculating the Inertion.

I am then having problems reading the steel charts to identify what I-beam would be sufficient.


Thanks for your persistence guys.




 

RE: Identificaiton of Beam requirements

Roof :
Total surface area:  4.5*1.6=7.2m^2  why 1.6?  It must be more than 2.13
SA attributed to beam:  7.2m^2/2 = 3.6m^2

Floor:
Total surface area:  4.5*2.13  = 9.6m^2
SA attributed to beam: 9.6m^2/2 = 4.8m^2
Wall:
Total Surface Area:  = 9.6m^2

Joists = 5.3m^2 / 2 = 2.65m^2

DEAD LOADS
Roof : 766.08N/m^2 8 3.6m^2 = 2.758kN
Floor:  167.58N/m^2 * 4.8m^2 = 0.85kN
Wall : 287.287 N/m^2 * 9.6m^2 = 2.75kN
Joists: 301.64 * 2.65m^2 = 0.8kN

Unit Dead Loads (kPa)
Roof:   0.77 * 1.3 = 1.0 (assuming 40 degree slope)
Floor:  0.7 (to account for joists, drywall, electrical, etc.
Wall:   0.5 (glass can be pretty heavy)
Joists:  included in floor DL

LIVE LOADS
Wind: 430.92N/m2 * 3.6m^2 = 1.55kN
Snow: 700N/m^2 * 3.6m^2 = 2.52kN
Floor Live Load: 1900N/m^2 * 4.8 = 9.12kN (I am not sure what this is but I have added it  under your instruction – I am assuming it is for furniture / people etc)

Unit Live Loads (kPa)
Wind:  not considered
Snow:  1.0 (minimum snow load in code)
Floor: 1.9  (this is a building code value for residential occupancy in my code...it could be different in yours)


Dead Load Safety Factor = 1.4  
Live Load Safety Factor = 1.6
1.25 and 1.5 respectively in my code

The parts I am unsure of are:

If I am using the correct Youngs Modulus value, and the correct strength value.  200,000 MPa

If I use the UDL with the safety factor when calculating the Inertion.  (what is inertion?)

I am then having problems reading the steel charts to identify what I-beam would be sufficient.  My steel handbook lists all of the beams with their moment capacities for various values of unbraced length.

The calculation in blue is more or less the way I would record calculations for this beam.  


Beam Design
Span 4.5m;  tributary width = 2.13/2 = 1.07m
wd = 1.07(1.0 + 0.7) + 2*0.5 (wall) = 2.82 kN/m
wl = 1.07(1.0 + 1.9)................= 3.10

wt (total load per meter) ..........= 5.92
wf = 1.4*2.82 + 1.6*3.10............= 8.91 (factored load)

Mf = 8.91(4.5)^2/8 = 22.5 kN-m
Try W200x19 and check for deflection (I = 16.6e6 mm^4)
Mr = 58.1 when top flange is continuously supported
Mr = 21.9 when top flange braced at 4m centers, so make sure you brace at midspan (I would go for 1.5m spacing of braces)
Delta (total) = 5/384 * 5.92(4500)^4/(200000*16.6e6) = 9.5mm
Total deflection = L/472 which is acceptable.


       
 

BA

RE: Identificaiton of Beam requirements

I think civilperson is as frustrated as you :)

I will print the above Monday to have a closer look. At the meantime, can you upload a simple floor plan and elevation (scanned hand sketch will do). Also, please list and indicate the material for each element.

RE: Identificaiton of Beam requirements

(OP)
Hi Guys, thanks for this it really is appreciated, (much to my wife's horror as I have spent way to much time at the desk over the weekend!).

BAretired.  Fantastic, thank you.  In answer to your questions, the roof was a mistake on my part (once again) I took the dimensions of one side, hence I did not have to divide it by two. "Inertion" should be "inertia"  = I.

This is where I was getting confused, working out "I", with a maximum Deflection of 4%.  Thanks again.  The safety factors are from BS5400 (if my memory serves me correctly)

Kslee100, I have drawn a basic layout including materials.



Apologies if it is a bit rough.

  

RE: Identificaiton of Beam requirements

qems,

How do you put the picture in with the text?  The only way I know how to do it is to upload the file which can then be downloaded by the reader.  

This seems like a much better way.  Since joining EngTips, I have wondered about this.  Is it a trade secret?

BA

RE: Identificaiton of Beam requirements

BA:

Pay up to get the trade secret out :)
Everybody knows something better.

RE: Identificaiton of Beam requirements

(OP)
Hi not a problem, you have to store the image online somewhere. I uploaded it to our business site, then just type the following.

img http://www.qems.biz/daphnia/house/SUMMERHOUSE 002.jpg]

I have missed out the first [ which should go before the "img" to make the text show up in the post. ie. the statement

img http://www.qems.biz/daphnia/house/SUMMERHOUSE 002.jpg

should be enclosed in square brackets []

Here is a good page which is linked to the site that give some handy hints on formatting. http://www.tipmaster.com/includes/tgmlinfo.cfm?w=450&h=450

Not sure if I explained that clear enough, but try it out and have a look at the link, there is some really use tips.

Rob


 

RE: Identificaiton of Beam requirements

Rob,

Sorry to be so thick, but how do I store an image online somewhere?

BA

RE: Identificaiton of Beam requirements

gems:

Can you follow through on BA's cal? He's pinned everything down, except the roof slope (40 assumed, 27 frm sketch). However, it would have been well covered by the safety factors.

Note, you will need to check wind induced uplift on the floor deck/joist, and the roof, to ensure proper tie down.  

RE: Identificaiton of Beam requirements

(OP)
Hi BAretired

There are two ways that you can do this

The easiest way is if you own a website.  Simply upload the photo and then take note of the url that the phot appears online and place it in the [img] tags.

If you do not have this ability  then you can use sites like

http://www.photobucket.com

Register for an account and upload your photo.

Then click on the "blogger icon" (the orange one) below the photo next to the word "share".

Next click on the

"Get Link Code"



Next - copy the code in the second box down "Direct Link for layout pages"

Then simply paste this code into the [img ] tags in your post.

Seems complicated but once youh ave done it a few times it will only take a second.  If only Beam calcs were this easy winky smile

Hope this helps.

Rob
 

RE: Identificaiton of Beam requirements

(OP)
Hi Kslee1000

Thanks for that, yes I should be able to follow through with BA's  calculations.  I have spoken to my contractor today and he has suggested using the UK equivalent to a  flanged W6 x 20.

As I am using  Uk based standards, (for example a UB203 x 127) is there link to a table that includes the "I" values and Mr values.  That way I can do teh same calcs for a W6 x 20 equivalent.

Thanks again

Rob
 

RE: Identificaiton of Beam requirements

Rob,

The shape which I suggested was a W200x19 (metric designation) which is equivalent to a W8x13 (Imperial designation).  This was in keeping with your proposal to use a Universal beam, which I believe is similar to a Wide Flange.

Its properties are:
d = 203 mm
b = 102 mm
tf = 6.5 mm (flange thickness)
w = 5.8 mm (web thickness)
Ix = 16.6 x 10^6 mm^4
Sx = 163 x 10^3 mm^3
Zx = 187 x 10^3 mm^3

We have a W6x20 (Imperial) in Canada which is stronger and slightly stiffer than the W8x13.

You might want to mention to your contractor that a pair of channels is another option.  They would be a bit heavier  but you would install one each side of the existing wood beam and bolt through thereby eliminating the need to shore the existing structure.

   

BA

RE: Identificaiton of Beam requirements

You might want to mention to your contractor that a pair of channels is another option.  They would be a bit heavier  but you would install one each side of the existing wood beam and bolt through thereby eliminating the need to shore the existing structure.


--

Definitely like the bolt-on-a-pair-of-double-channels solution:  

You add strength and rigidity while NOT needing to shore up the whole structure while pulling and removing the old wood beam.  

Shoring can be done, but ANY problem or slipping threatens the whole building since replacement of the original beam requires (by definition) removing ALL the original structural strength BEFORE being able to put back the new beam.    Too many chances for something to go wrong and destroy your building.  Paint your new channels: use a good primer plus two cover coats.

Bolt one end of the new channels to the current beam, jack up the other end of the two channels and bolt them in position, making (as said) a sandwhich with the old wood structure in the middle of the two channels: You will almost certainly see the "sag" in the middle in the current wood beam.   

Then press up the middle of the old sagging beam until it matches the "straight" new channels.  Drill your new holes through the channels and through the old truss (at the bottom) - no smaller than 1/2 dia (12 or 13 mm) high grade bolts every 150-200 mm - and clamp firmly.    

Result will be a safer, easier installation; prettier than a WF that will be stronger as well.

RE: Identificaiton of Beam requirements

(OP)
HI Guys

Kslee1000: Thanks for the charts, excellent, and greatly appreciated.

BAretired:

The W6 x 20 is the equivalent of what  he is wanting to use, looking at the charts.

As for the pair of channels, all of the joists are rotten and having to be removed along with the floor, the only original part of the structure that will remain will be the roof. Hence we are not worried about having to shore, as it all has to come out anyway.

How did you get on with the image inserting?

Rob
   

RE: Identificaiton of Beam requirements

Rob,

If the floor is rotten, then I agree with you.  Use the Universal beam.

As for the imaging, I don't have my own web site.  I have successfully loaded an image into photobucket, but so far I haven't had any occasion to send it to anyone.  One concern is that I seem to be getting a lot spam popping up from that site.  I also have some concerns as to the safety of the site as everyone and his dog seems to have access to everyone else's photos.  I will mull it over for a while, but thanks for the info.  At least I know how to do it now.

BA

RE: Identificaiton of Beam requirements

(OP)
Hi Racookpe1978

Thanks for that.

The only load on the old beam now is the existing floor, which is coming out.  see below:



The scaffolding you see going through the floor is load bearing and taking the weight of the roof on teh Right hand side.  The other side is anchored onto the existinghosue wall.  



The hopefully simple plan is that the contractor removes the floor and Joists.  Inserts the Upright column from ground to Roof, bolts up the new beams and re-instates the floor.  Then it is a case of putting in the intermediate supports and filling the blank spaces. winky smile  All steelwork will be covered with hardwood as per my wifes specification! winky smile

If it goes up as easily as it came down I will be happy.  That said, we have been doing this house up for 7 years and it always throws little surprises at us to make life interesting!

Thanks again.

rob

 

RE: Identificaiton of Beam requirements

(OP)
Hi BA

I am not sure, but I think you can set your preferences on the site so that no-one else can see your images.  That said if you are putting them in a forum then they are in the public domain anyway.  

When you register, so long as you didn't tick the boxes saying send me info on..... etc, you should be ok for spam.  

Failing that, if you ever need images posted, drop me a line and I can upload them to our server for you.  It is the least I could do for all your help.

Rob
 

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